Answer:
I'm guessing it's C because it's the only one that looks right
The two friends will be 150 miles apart, when the sum of the distances covered by the them is 150 miles.
distance = speed x time.
Let the time covered when they are 150 miles apart be x, then:
40x + 35x = 150
75x = 150
x = 150/75 = 2.
Therefore, the two friends will be 150 miles apart in 2 hours.
Answer:
![\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3D%5Ctext%7B%2012%7D%5Cdegree%20%5C%5C%20B%3D%5Ctext%7B%20114%7D%5Cdegree%20%5C%5C%20C%3D54%5Cdegree%20%5Cend%7Bgathered%7D)
Step-by-step explanation:
To calculate the angles of the given triangle, we can use the law of cosines:
![\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ccos%20%28C%29%3D%5Cfrac%7Ba%5E2%2Bb%5E2-c%5E2%7D%7B2ab%7D%20%5C%5C%20%5Ccos%20%28A%29%3D%5Cfrac%7Bb%5E2%2Bc%5E2-a%5E2%7D%7B2bc%7D%20%5C%5C%20%5Ccos%20%28B%29%3D%5Cfrac%7Bc%5E2%2Ba%5E2-b%5E2%7D%7B2ca%7D%20%5Cend%7Bgathered%7D)
Then, given the sides a=2, b=9, and c=8.
![\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ccos%20%28A%29%3D%5Cfrac%7B9%5E2%2B8%5E2-2%5E2%7D%7B2%5Ccdot9%5Ccdot8%7D%20%5C%5C%20%5Ccos%20%28A%29%3D%5Cfrac%7B141%7D%7B144%7D%20%5C%5C%20A%3D%5Ccos%20%5E%7B-1%7D%28%5Cfrac%7B141%7D%7B144%7D%29%20%5C%5C%20A%3D11.7%20%5C%5C%20%5Ctext%7B%20Rounding%20to%20the%20nearest%20degree%3A%7D%20%5C%5C%20A%3D12%C2%BA%20%5Cend%7Bgathered%7D)
For B:
![\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ccos%20%28B%29%3D%5Cfrac%7B8%5E2%2B2%5E2-9%5E2%7D%7B2%5Ccdot8%5Ccdot2%7D%20%5C%5C%20%5Ccos%20%28B%29%3D%5Cfrac%7B13%7D%7B32%7D%20%5C%5C%20B%3D%5Ccos%20%5E%7B-1%7D%28%5Cfrac%7B13%7D%7B32%7D%29%20%5C%5C%20B%3D113.9%5Cdegree%20%5C%5C%20%5Ctext%7BRounding%3A%7D%20%5C%5C%20B%3D114%5Cdegree%20%5Cend%7Bgathered%7D)
Step-by-step explanation:
Fractions with only factors of 2 and 5 are terminating. If not, they give repeating decimals.
Therefore 17/8 and 34/16 are terminating whereas 2/13, 5/24 and 6/7 are repeating decimals.
1/3 ln(<em>x</em>) + ln(2) - ln(3) = 3
Recall that
, so
ln(<em>x</em> ¹ʹ³) + ln(2) - ln(3) = 3
Condense the left side by using sum and difference properties of logarithms:
![\log_b(m)+\log_b(n)=\log_b(mn)](https://tex.z-dn.net/?f=%5Clog_b%28m%29%2B%5Clog_b%28n%29%3D%5Clog_b%28mn%29)
![\log_b(m)-\log_b(n)=\log_b\left(\dfrac mn\right)](https://tex.z-dn.net/?f=%5Clog_b%28m%29-%5Clog_b%28n%29%3D%5Clog_b%5Cleft%28%5Cdfrac%20mn%5Cright%29)
Then
ln(2/3 <em>x</em> ¹ʹ³) = 3
Take the exponential of both sides; that is, write both sides as powers of the constant <em>e</em>. (I'm using exp(<em>x</em>) = <em>e</em> ˣ so I can write it all in one line.)
exp(ln(2/3 <em>x</em> ¹ʹ³)) = exp(3)
Now exp(ln(<em>x</em>)) = <em>x </em>for all <em>x</em>, so this simplifies to
2/3 <em>x</em> ¹ʹ³ = exp(3)
Now solve for <em>x</em>. Multiply both sides by 3/2 :
3/2 × 2/3 <em>x</em> ¹ʹ³ = 3/2 exp(3)
<em>x</em> ¹ʹ³ = 3/2 exp(3)
Raise both sides to the power of 3:
(<em>x</em> ¹ʹ³)³ = (3/2 exp(3))³
<em>x</em> = 3³/2³ exp(3×3)
<em>x</em> = 27/8 exp(9)
which is the same as
<em>x</em> = 27/8 <em>e</em> ⁹